Question: Find the limit as $x$ approaches positive infinity. $\lim_{x\to\infty}\dfrac{\sqrt{9x^6+4x^2}}{x^3-1}=$
Solution: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the denominator is $x^3$, let's divide by $x^3$. In the numerator, let's divide by $\sqrt{x^6}$, since for positive values, $x^3=\sqrt{x^6}$. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{\sqrt{9x^6+4x^2}}{x^3-1} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{\sqrt{9x^6+4x^2}}{\sqrt{x^6}}}{\dfrac{x^3-1}{x^3}} \gray{\text{Divide sides by }x^3=\sqrt{x^6}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to\infty}\dfrac{\sqrt{\dfrac{9\cancel{x^6}}{\cancel{x^6}}+\dfrac{4\cancel{x^2}}{\cancel{x^2}\cdot x^4}}}{\dfrac{1\cancel{x^3}}{\cancel{x^3}}-\dfrac{1}{x^3}} \\\\ &=\lim_{x\to\infty}\dfrac{\sqrt{9+\dfrac{4}{x^4}}}{1-\dfrac{1}{x^3}} \\\\ &=\lim_{x\to\infty}\dfrac{\sqrt{9+0}}{1-0} \gray{\lim_{x\to\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{\sqrt{9}}{1} \\\\ &=\dfrac{3}{1} \\\\ &=3 \end{aligned}$ In conclusion, $\lim_{x\to\infty}\dfrac{\sqrt{9x^6+4x^2}}{x^3-1}=3$.